# Maximum likelihood estimator of the exponential distribution with two same parameter [duplicate]

Let $X_{1},...,X_{n}$ be a random samples from a population having the probability density function:

$f(x)= \frac{1}{\theta}e^{\frac{-(x-\theta)}{\theta}}$ where $x\geq \theta$.

When I solved the likelihood equation of the above density, the maximum likelihood estimator turned out to be mean of the sample but if you see the range of values that the parameter $\theta$ can take is $(0,X_{1}]$. How can this be possible that the likelihood estimator is out of the range of $\theta$. Please clear my doubt regarding this. Thanks

• In the standard approach to deriving the maximum likelihood estimator of a parameter vector, it's necessary to assume that the support of the distribution of the data does not depend on the parameter(s). This assumption is violated for the exponential distribution. That's why one has to pursue a different route altogether. – Mico Feb 3 '18 at 14:58
• Why Duplicate? In that question, I asked about method of moment estimator. But this question is about maximum likelihood estimator. Only density is same. Before marking it as duplicate, at least the referred question should be checked whether it is actually duplicate or not. – user8125394 Feb 3 '18 at 16:37

I suspect you have solved the maximum likelihood estimation problem incorrectly, most likely by ignoring the parameter constraint. From the stated density, your log-likelihood function is:

\begin{aligned} l_\boldsymbol{x}(\theta) &= -n \ln \theta - \sum_{i=1}^n \frac{x_i-\theta}{\theta} \\[8pt] &=-n \ln \theta - \frac{n \bar{x}_n}{\theta} + n \\[8pt] &= n \left( 1 - \ln \theta -\frac{\bar{x}_n}{\theta} \right) \text{ } \text{ } \text{ } \text{ } \text{ for all } 0 < \theta \leqslant \min (x_1 ,..., x_n). \end{aligned}

For all $0 < \theta < \min (x_1 ,..., x_n)$ this has first derivative:

$$\frac{dl_\boldsymbol{x}}{d\theta}(\theta) = n \left( - \frac{1}{\theta} + \frac{\bar{x}_n}{\theta^2} \right) = \frac{n}{\theta^2} (\bar{x}_n - \theta).$$

Now, over the range $0 < \theta < \min (x_1 ,..., x_n)$ we have $\theta < \bar{x}_n$ which means that the first derivative is strictly positive. Hence, the maximising value occurs at the border case:

$$\hat{\theta} = \min (x_1 ,..., x_n).$$

I suspect that you have probably ignored the parameter constraint and solved the score equation (setting the derivative of the log-likelihood to zero). That would yield the incorrect solution of the sample mean.